# Find the partial pressure in a container

I have an air mass which consists of 78.1% $\ce{N_2}$, 20.9% $\ce{O_2}$, 0.964% $\ce{Ar}$, and 0.0378% $\ce{CO_2}$ in a container which volume $\mathrm{V=40~L}$ at a temperature $\mathrm {T = 298.15~K}$.

The total pressure is $\mathrm{p_{total} = 1.5~atm}$.

Now I am removing all the $\ce{O_2}$ in the container. What will the partial pressure of $\ce{N_2}$ be?

I know the formula $\mathrm{pV = nRT}$.

Before I remove all the $\ce{O_2}$ I can calculate the amount of substance $$\mathrm{{n_{total} = \frac{pV}{RT} = \frac{1.5 \cdot 40}{0.082 \cdot 298.15}~mol = 2.45~mol}}$$.

The molar mass of $\ce{N_2}$ and $\ce{O_2}$ is $\ce{M(N_2)} = \mathrm{28~g\ mol^{-1}}$ and $\ce{M(O_2)} = \mathrm{32~g\ mol^{-1}}$.

If I remove all the $O_2$ the new mass/volume percentage of $\ce{N_2}$ will be

$$\mathrm{m_\%(N_2) = \frac{78.1}{78.1+0.964+0.0378} = 98.7 \%}$$. So I thought I could find the amount of substance of $\ce{N_2}$ by saying $$\ce{n(N_2)} = \mathrm{\frac{98.7~g}{28~g/mol} = 3.525~mol}$$ but this will exceed the total amount of substance before I removed $\ce{O_2}$ so this makes no sense.

When I have found the amount of substance of $\ce{N_2}$ I think I can just replace $\ce{n}$ in $\mathrm{p = \frac{nRT}{V}}$ and get the partial pressure of $\ce{N_2}$.

Maybe I have the right percentage of $\ce{N_2}$ but I am calculating the amount of substance wrong when it's a gas? Could the correct answer be $\mathrm{98.7 \% \cdot n_{total} = 2.422~mol}$ and then inserting this in $\mathrm{p = \frac{nRT}{V}}$ which gives me around $\ce{p(N_2)} = \mathrm{1.5~atm}$?

Since we're assuming these are all ideal gases, we can use Dalton's Law of Partial Pressures to solve this fairly easily; you never need to find moles of anything. Unless something is strange, the percentages given are by volume or by pressure rather than by mass. If they are intended to be by mass, what follows would need to be modified. I suspect that the use of the words "air mass" in the problem are what caused the confusion.

The total pressure, 1.5atm, is equal to the sum of the partial pressures, each of which is proportional to its mole fraction in the container. Because these are gases that have the same volume, 40L, their mole fractions are their percent of the total amount of gas.

The nitrogen, which is 78.1% of the total, has a partial pressure of:

$1.5 \cdot 0.781 = 1.1715\ atm$

That's true whether the oxygen is in the container or not. To see that this is the case, we can remove the oxygen and do the same calculation.

If we remove 20.9% of the gas (in the form of oxygen) we are left with 81.1% of the original 1.5 atm remaining.

$1.5\cdot 0.811 = 1.1865\ atm$

We can determine the mole fraction of nitrogen in this new mixture of gases by adding up the remaining percentages and finding nitrogen's contribution to it:

$78.1 + 0.964 + 0.0378 = 79.1018\ units\ of\ gas\ total$

$\frac{78.1}{79.1018} \cdot 100\% = 98.73\%\ N_2\ after\ oxygen\ is\ removed$

As before, Dalton's Law of Partial Pressures tells us that the pressure of nitrogen will be its mole fraction times the total pressure, or 98.73% of 1.1865 atm.

$1.1865 \cdot 0.9873 = 1.1714\ atm$